The Triangle Game

Question

Take an equilateral triangle with dots at the vertices and at the midpoints of each side:

There are a total of six dots.

In the triangle game, you have the numbers 1, 2, 3, 4, 5, 6. Place one number at each dot such that the numbers on each side adds up to the same total.

Here's a solution for side length 9:

Find all the distinct solutions of the triangle game, not including reflections and rotations.

Answer 1

Let $a,b,c$ be the numbers at the corners and $d,e,f$ be the numbers at the midpoints (say $d$ opposite $a$, $e$ opposite $b$, $f$ opposite $c$). To fix the rotation and reflection of a given solution, we can assume $a\leq b\leq c$. Clearly the largest of the three midpoint-numbers must be opposite the largest of the three corner-numbers and so on, so we also have $d\leq e\leq f$.

Let $x=a+b+c$ and $y=d+e+f$, and let $S$ be the sum from each side. We have two equations in $x$ and $y$:

• $x+y=21$ (the sum of $1,\dots,6$)
• $2x+y=3S$ (adding together all three copies of $S$ results in counting each corner twice and each midpoint once).

So $x$ and $y$ must both be multiples of $3$. Thus the three numbers $a,b,c$ summing to $x$ must consist of one that is a multiple of 3, one that is one more than a multiple of 3, and one that is one less that a multiple of 3; similarly for $d,e,f$. So the possibilities are:

• $a=1,b=2,c=3,d=4,e=5,f=6$, giving $S=9$
• a=1,b=2,c=6,d=3,e=4,f=5, required condition not satisfied
• $a=1,b=3,c=5,d=2,e=4,f=6$, giving $S=10$
• a=1,b=5,c=6,d=2,e=3,f=4, required condition not satisfied
• $a=4,b=5,c=6,d=1,e=2,f=3$, giving $S=12$
• a=3,b=4,c=5,d=1,e=2,f=6, required condition not satisfied
• $a=2,b=4,c=6,d=1,e=3,f=5$, giving $S=11$
• a=2,b=3,c=4,d=1,e=5,f=6, required condition not satisfied